2 Answers
2

From this we have: $\bigcup x^+ = \{z\mid\exists y\in x\cup\{x\}: z\in y\}$. We then follow the definition of a transitive set:

The set $x$ is transitive if for every $y\in x$, $y\subseteq x$

Now we want to show that $x\subseteq\bigcup x^+$ and that $\bigcup x^+\subseteq x$, the axiom of extensionality will ensure equality.

For the first one, it is nearly trivial. Since $x\in x^+$ we have that $y\in x$ then $y\in\bigcup x^+$ immediately (recall the definition of this union).

For the second one, we recall that $y\in x^+$ then $y\in x$ or $y=x$. Therefore $y\subseteq x$. So for every $z\in\bigcup x^+$ we have some $y\subseteq x$ such that $z\in y$, therefore $z\in x$ as wanted.

The set $x$ is transitive if $z\in y \in x$ implies $z\in x$. Moreover, $x^+=x\cup\{x\}$. So let $x$ be a transitive set.

We have $x\subseteq\bigcup x^+$ since $x\in\{x\}$. Now let $z\in \bigcup x^+$. Then $z\in\bigcup x\cup x$. If $z\in\bigcup x$, then there exists $y\in x$ with $z\in y$. But since $x$ is transitive, $y\subseteq x$ and hence $z\in x$.